TSPSC AEE 2022 EE Questions-with-Solutions Updated
TSPSC AEE 2022 EE Questions-with-Solutions Updated
TSPSC AEE 2022 EE Questions-with-Solutions Updated
Questions With
Detailed Solutions
ELECTRICAL ENGINEERING
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SUBJECTWISE WEIGHTAGE
No. of
S.No. Name of the Subject
Questions
1 EMT 9
2 Electric Circuits 16
3 Control Systems 11
4 Electrical Machines 35
5 Power Systems 15
6 Electrical & Electronic Measurements 6
7 Power Electronics & Drives 23
8 Signals and Systems 5
9 Analog Electronics 14
10 Digital Electronics 8
11 UEE 8
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01. Consider an infinite uniform line charge of 5 nC/m v0(t) = Instantaneous output voltage of 3-f FW
along the x axis in free space. Calculate the electric uncontrolled rectifier.
field at a distance of 2m from the line charge along v0(t) = vA0 − vB0 = line voltage
the z axis. Peak or maximum value of v0(t) = Vml
(a) 45 zt V/m (b) 90 zt V/m = 2 # V, rms
(c) 22.5 zt V/m (b) 11.25 zt V/m = 2 # 400 V
01. Ans: (a) = 400 2 V
Sol: x 03. In a single-phase transformer supplied from a
constant input voltage, the magnitude of the load
5nC/m (0,0,2) current is kept constant while the power factor is
z
varied. Under this condition
y (a) the maximum efficiency occurs at unity power
factor
, t ,
E
> H (b) the power factor where maximum efficiency
2 o 2 o 2
occurs depends on the leakage inductance
5 # 10 9 2zt
= = 2G values
1 ]2g (c) the maximum efficiency occurs at power
2 #
36 # 10 9
factor of 0.5(lead)
5 # 18 (d) the maximum efficiency occurs at power
= zt = 45zt V/m
2 factor of 0.5 (lag)
03. Ans: (a)
02. A three-phase diode bridge rectifier supplied from a Sol: Efficiency is maximum at UPF.
three-phase, 400V, 50Hz ac supply delivers power
04. A synchronous generator rated 11kV, 50MVA has a
to a resistive load of 50Ω. The peak value of the
per unit impedance of 0.2pu on its own base. Then
instantaneous load voltage would be
its impedance referred to a 22kV, 150MVA base
(a) 400 ]2/3g (b) 400/ 3
would be
(c) 400 2 V (d) 400V
(a) 0.15pu (b) 0.133pu
02. Ans: (c)
(c) 0.2pu (d) 0.1pu
Sol:
n0(t) 04. Ans: (a)
.e o
S B(new) VB(old) 2
Sol: X pu(new) = X pu(old)
Vml S B(old) VB(new)
wt
X pu(new) = 0.2 # b 50 lb 22 l = 0.15 pu
0° 30° 90° 150 11 2
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10
05. Consider the following statements about a linear Steady state value = Lts→0 s x (s) = =2
5
system that has a transfer function given as
1s
G]s g
07. Two synchronous generators G1 and G2 rated
:
1s 200MW and 400MW respectively are operated
A. G(s) is a minimum-phase system in parallel to supply a total load of 300MW. If
B. The system is BIBO stable the governors in both the machines are set to a
C. It is an all pass system droop of 4%, what would be the individual power
Which of the above statements is/are true? supplied by each generator?
(a) A, B and C (b) B only (a) G1 = 100 MW, G2 = 200MW
(c) A and B only (d) B and C only (b) G1 = 50 MW, G2 = 250MW
05. Ans: (d) (c) G1 = 200 MW, G2 = 100MW
1s (d) G1 = 150 MW, G2 = 150MW
Sol: G(s) =
1s 07. Ans: (a)
Sol:
s = −1 s = +1
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X +5V
V0 V(t)
2 kΩ Y
+5V
1 kΩ
FWR PMMC
2V +
−
-5V V0 avg
0A +
V0
0A −
2 kΩ 2V0 + 2 1 kΩ
2V +
− 3
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16. Find Vout in the below circuit. Assume opamp to be To increase the range of an ammeter, the shunt
ideal. resistance is used.
1 kΩ Rm 0. 0.1
Rsh = = = 0.05 W
d 1n d 1n
I 30 2
1 kΩ Im 10
+
−
1V 18. A MOSFET biased in common-drain configuration
+ Vout
− 1 kΩ is best suited for designing a
(a) transresistance amplifier
(a) −1V (b) −2V (b) transconductance amplifier
(c) 0V (d) 1V (c) current buffer
16. Ans: (a) (d) voltage buffer
Sol: 1 kΩ 18. Ans: (d)
+VDD
Sol:
+ 0.5 −
I
1 kΩ I i/p
+
(0V) − o/p
1V (0V)
+ Vout
−
1 kΩ MOSFET in CD (common drain) source
follower
I=
1V
= 0.5 mA Voltage gain Av ≅ 1
2
∴ Voltage buffer
V0 = 0.5 mA × 1kΩ = − 0.5 V
Nearest option (a) is correct 19. What is the condition to achieve regenerative
braking of induction motor?
17. An ammeter has a range of 0 − 10A with an internal
(a) synchronous speed should be doubled
resistance of 0.1Ω. In order to increase its range to
(b) synchronous speed should be increased by a
0 − 30A, we need to add a resistance of
factor of 1.5
(a) 0.05Ω in shunt with the meter
(c) synchronous speed should be a little higher
(b) 0.1Ω in shunt with the meter
than the rotor speed
(c) 0.05Ω in series with the meter
(d) synchronous speed should be a little lower
(d) 0.1Ω in series with the meter
than the rotor speed
17. Ans: (a)
19. Ans: (d)
Sol: Im = 10 A, Rm = 0.1 W
Sol: For regenerative mode N > Ns
I = 30 A; Rsh = ?
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20. What is the low order ripple frequency of the 22. What is the unit of illumination?
output voltage of a three phase fully controlled (a) coulomb (b) lux
bridge converter, if the AC input supply frequency (c) decibel (d) henry
is f ? 22. Ans: (b)
(a) 3f (b) 6f Sol: The unit of illumination is lux
(c) f (d) 2f
20. Ans: (b) 23. Which bridge is used to measure frequency?
Sol: In 3-f fully controlled bridge converter, the output (a) Wien’s bridge (b) Anderson bridge
voltage wave form consists of six pulses for every (c) Maxwell’s bridge (d) Schering bridge
one cycle of input voltage wave form then it is also 23. Ans: (a)
called 3-f six pulse controlled converter. Sol: Wien’s bridge is usd to measure the frequency.
f0 = 6 f.
24. A constant V/f controlled induction motor is
21. Which of the following armature voltage control fed from a variable voltage variable frequency
method is employed when the supply is dc? three phase voltage source inverter. The motor
(a) Chopper control is operated within the base speed. Which of the
(b) Static Ward - Leonard schemes following is true in torque-speed characteristics of
(c) Ward-Leonard schemes this motor?
(d) Transformers with taps and an uncontrolled (a) starting torque increases with decrease in
rectifier bridge frequency and maximum torque decreases
21. Ans: (a) (b) starting torque decreases with decrease in
frequency and maximum torque decreases
Sol:
(c) starting torque increases with decrease in
+ + Ia frequency and maximum torque remains
Vdc Chopper Vt unchanged
Eb ra
- - (d) starting torque decreases with decrease in
frequency and maximum torque remains
Vt I a ra unchanged.
N ∝ 24. Ans: (c)
Sol: Starting torque Tst ∝ d n
1 V 2
f f
Fo the speed control of DC motor, it input supply is
Maximum torque Tmax ∝ d n
dc then the chopper converter is used in armature V 2
f
voltage (Vt) control method. V
With constant , maximum torque is constant
f
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and starting torque is inversely proportional to 26. Two channels of a CRO are fed with two signals.
frequency. In the X-Y mode, an ellipse with major axis
aligned along the Y axis is observed. The following
25. Q1 and Q2 are perfectly matched BJTs. Assuming inference can be made from this
beta to be infinite and forward bias voltage drop in (a) Two signals are periodic with same amplitude
diode to be 0.7V, find the current Iout. but different frequency
+5V (b) Two signals are periodic with same frequency
but different phase and amplitude
Q1 Q2 (c) Two signals are periodic with same frequency
and amplitude but different phase
(d) Two signals are periodic with same frequency
1 kΩ Iout and phase but different amplitude
−5V 26. Ans: (b)
Sol: Y
(a) 4.3mA (b) 5.7mA →f =90°,fy = fx, Ay = Ax
(c) 0mA (d) 3.6mA
25. Ans: (d) X
Sol: +5V
Y
+0
X
Ix Iout
+
0.7 −5V
− Y
1 kΩ
→f =90°, fy = fx, Ay > Ax
β⇒∞ X
IB → 0
In the X-Y mode, an ellipse with major axis aligned
Current mirror along the Y axis is obsorbed then the two signals
5 − 0.7 − 0.7
Iout = Ix = are periodic with same frequency but different
1k
3.6 phase and amplitude.
= = 3.6 mA
1k
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27. A single-phase diode bridge rectifier is used to 29. Consider the following lists regarding
supply a highly inductive load. If the load current compensation techniques in power systems:
is assumed to be ripple free, then the input current A. Reduce the Ferranti effect
at the ac side of the rectifier will be B. Improve power factor
(a) Triangular wave C. Increase power flow capability of line
(b) Square wave D. Reduce current ripple
(c) Purely sinusoidal 1. Series capacitor
(d) Pure DC 2. Shunt reactor
27. Ans: (b) 3. Shunt capacitor
Sol: 1-f FWR, 4. Series reactor
Choose the option in which all items are correctly
I0 matched.
(a) A−4, B−2, C−1, D−3
wt (b) A−2, B−3, C−1, D−4
Is
(c) A−2, B−1, C−3, D−4
p+a (d) A−1, B−3, C−2, D−4
2p+a wt
a 29. Ans: (b)
Sol: Shunt Capacitor → Improves Power factor
In a single phase diode bridge rectifier (1-f FWR) Series Capacitor → Improves Power flow
is used to supply a highly inductive load. If load capability
current (I0) is assumed to be ripple free (I0= Shunt reactor → reduces the ferranti Effect
constant) then the input current at the AC side of Series reactor → reduces the current ripples
the rectifier will be square wave (Is).
30. The surge impedance of a 300 km long overhead line
28. A three-phase 33kV, oil-circuit breaker is rated is 180 ohms. For a 150 km length of the same line,
1500A, 2000MVA, 2s. The symmetrical breaking the surge impedance in ohms would be
current for this breaker would be (a) 90 Ohms (b) 270 Ohms
(a) 50 kA (b) 40 kA (c) 180 Ohms (d) 360 Ohms
(c) 25 kA (d) 35 kA 30. Ans: (c)
28. Ans: (d) Sol: Surge impedance never depends on the length of
Sol: The symmetrical breaking capacity = 3 VL # Isym the line.
= 2000 # 106
=
` Isym 34.99kA
3 # 33 # 103
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rotating at synchronous speed, but the rotor is at Sol: MOSFET is unipolar, unidirectional switch.
stationary, therefore the rotor field (or poles or MMF) Diode is unipolar, unidirectional switch.
will be at zero speed. The relative speed between SCR is Bipolar, unidirectional switch.
stator MMF & rotor MMF is equal to synchronous BJT is unipolar, unidirectional switch.
speed. But for the production of unidirectional torque Voltage bidirectional two quadrant switch means
both must be stationary with respect to each other. bipolar, unidirectional switch. SCR is bipolar
unidirectional switch.
36. Consider the following statements about three-phase i
voltage source inverters:
A. they require voltage-bidirectional two quadrant
v
devices to realise the switches
B. PWM techniques are used to reduce the
frequency of the harmonics in the output
C. in the 180o conduction mode, a new switch is 38. What should be the value of the capacitance so
gated in every 60 duration.
o
that the resonance frequency of the circuit below is
Which of the above statements is/are true? 314.16 rad/sec?
(a) A only (b) C only
(c) A and B only (d) B and C only 1 Ohm
+
− 2 Ohms
36. Ans: (d) 5V C
Sol: • Voltage source inverter (VSI) require unipolar 0.1 H
and bidirectional switch like MOSFET, IGBT
with body diodes. (a) C = 0.1 mF (b) C = 0.01 mF
• Objectives of PWM techniques are to control (c) C = 1 mF (d) C = 10mF
rms value of the output voltage and to eliminate 38. Ans: (a)
(or) minimise the lower order harmonic content Sol:
to this possible extent. 1W
5V +
− C 2W
• In the 180° conduction mode, a new switch is
0.1 H
gated in every 60° duration.
37. Which among the following is an example for a
voltage bidirectional two-quadrant switch? Y = Y 1 + Y2 + Y 3
(a) MOSFET (b) SCR
1 1 1
=
(R1 jX L) jX C R 2
(c) BJT (d) Diode
1 j 1
37. Ans: (b) = + +
(1 + jX L) X C 2
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1 jX L j 1
= 15.2W
1X XC 2
L 2
= e o je o
1 1 1 XL
2 1 X L2 X C 1 X L2 3A 7A
50W
190W
47.5W
For Resonance, imaginary part = 0
1 XL
⇒ 1 X 2L X C X L 39.6W
X C 1 X 2L
L By S.T T.
1 + (w0L)2 = ⇒ C = 0.1 mF 15.2W 39.6W
C V
A
20 Ohms
I2 10 Ohms 40 Ohms I1
By Nodal analysis at point, A
3A 30 Ohms 50 Ohms 7A
V ( 277) V
=3
15.2 39.6 47.5
(a) 26.17V (b) 2.617V
47.5V + 13157.5 + 54.8 V = 3 × 47.5 × 54.8
(c) 52.35 V (d) 5.235V − 5348.5
39. Ans: (c) V=
102.3
Sol: By star-delta transformation V = 52.28 V
20 W
s
63.33 W 40. Laplace transform of f(t) is . Then the f(t) is
s −4
2
3A 7A
50W
47
W
0
.5W
19
41. An infinitely long current filament carrying 28.1A
of current in the positive z direction. The magnetic
field intensity, H, at ^ 20 , 0, 4 h
(a) 0.1 yt A/m (b) 10 yt A/m
(c) 1 yt A/m (d) 0.5 yt A/m
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(0,0,z)
I.p.f = 0.9cosa
at , = at z
SP (0,0,4) The power factor at the ac input side depends on
at , at z the converter firing angle.
v
R The converter can be operated with firing angle
at=
= at=
R greater than 90° also. The range of firing angle is
v|
|R
0° to 180°.
20
at = = at x
20 43. A DC shunt motor with an armature resistance of
H due to infinite line is given by 0.15Ω is supplied from 230V input supply. If the
back emf of the motor is 200V, then the armature
^at , # at =h
I
H
2R current will be equal to
^at z # at xh (a) 150A (b) 250A
28.1
2# 20 (c) 100A (d) 200A
` H = at y A/m 43. Ans: (d)
V E b 230 200
Sol: Ia =
42. Consider the following statements about a phase Ra 0.15
controlled single-phase full-bridge converter using 30
= = 200 A
SCRs: 0.15
A. The average output voltage at the dc side
varies linearly with the firing angle 44. Consider the following statements about the
B. The power factor at the ac input side depends operation of a synchronous machine:
on the converter firing angle A. the armature reaction in the generating modes
C. The converter cannot be operated with firing aids the field flux when supplying a leading
angle greater than 90 degrees current
Which of the above statements is/are true? B. the armature reaction opposes the field flux
(a) B only (b) A only for a motor that draws a leading power factor
(c) A and B only (d) B and C only current
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S R conduction.
1
` PL \ 2 V0
D
V
0.6 #
50 75V
Vs 1 D s 1 0.6
46. The MOSFET in the circuit shown below is But given output voltage is 50V, so it is not
operated as a switch at a frequency of 10kHz continuous conduction mode. At the same time,
and duty ratio of 0.6. The initial inductor current it is not discontinuous conduction mode. (DCM)
is zero. If all the components can be assumed to because, in D.C.M output voltage will be more
be ideal, what would be the energy stored in the than 50V.
inductor at the end of 10 switching cycles?
So in every cycle inductor getting energy in 0.6 T
and delivering energy during 0.4 T duration. The
slope of current during ON time (TON) and OFF
time (TOFF) are as follows.
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22-01-2023
iL
(a) Xd > Xq (b) Xd = − Xq
3T (c) Xd < Xq (d) Xd = Xq
− 50 # t
50 # L
L t 47. Ans: (a)
300 mA
200 mA
Sol: Xd > Xq
100 mA
t
TON T 2T 48. Consider the following statements made about the
sequence impedance of power system components.
VS #
L t...........TON A. a fully transposed transmision line has equal
V0 # positive and negative sequence impedances
L t ........TOFF B. the negative sequence impedance of a
L = 10 mH synchronous generator is usually smaller than
1 the positive sequence impedance
T 100s
10 # 103 C. the negative sequence impedance of a
D = 0.6 transformer is generally much higher than the
During 1st cycle, at the end of TON, positive sequence impedance
50 Which of the above statements is/are true?
I L1 = # 100 # 10 -6 # 0.6 300mA
10 # 10 -3 (a) A and C only (b) A only
50
at the end of T; IL2 = 300 − L # TOFF (c) A and B only (d) B and C only
48. Ans: (c)
50
I L2 300 # 10 -3 # 0.4 # 100 # 10 -6 Sol: For generator
10 # 10 -3
X1 ≥ X2 >> X0
= 100 mA
For Transformer
At the end of 1st cycle it gets a current of 100 mA.
X1 = X2 = X0
It is continuous so at the end of 10th cycle, current
For Transmission Line
through inductor is 1000 mA. i.e., 1A
X1 = X2 << X0
1
∴ Stored energy in inductor = 2 LI2
= 2 # 10 # 10 -3 # (1) 2
1 49. A dynamometer type wattmeter is used to measure
power of a room heater. Which option is correct?
= 5 mJ
(a) Current in the fixed coil is lower than the
47. In a salient pole synchronous generator with Xd current in the moving coil
and Xq denoting the reactance in the d and q axes, (b) Cannot comment on the relative currents in
what is likely to be the relation between Xd and Xq? the moving and fixed coil
(c) Current in the fixed coil is same as the current
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V0 1 s2 C2 R2
∴ = 1 ...............(4) Power (MSCP).
Vin 1 s2 C2 R2
(a) 4000/π (b) 8000π
The given circuit is an All Pass Filter.
(c) 2000π (d) 4000π
54. Ans: (a)
52. The decimal equivalent of the HEX number EF.A
Sol: Mean spherical candle power (MSCP)
is
Total lu min ous flux
(a) 239.625 (b) 239.16 =
4
(c) 239.15 (d) 239.6
16000 4000
52. Ans: (a) = =
4
Sol: (EF.A) = 161 × E + 160 × F + 16−1 × A
= 224 + 15 + 0.625
= (239.625)10 55. What is the effect on co-efficient of adhesion due
to following conditions on rails?
53. A three-phase full controlled converter (with 6 (a) it is high when the rails have grease
SCRs only ) is feeding the armature of a separately (b) it is high when the rails are dry
excited DC motor. The motor has to also operate (c) it is high when the rails are oiled
in quadrant-III. Which of the following methods is (d) it is high when the rails are wet
suitable? 55. Ans: (b)
(a) By operating with triggering angle a > p/2 rad Sol: The coefficient of adhesion is high when the rails
(b) By connecting a free wheeling diode across are dry.
the armature in addition to adjusting the
triggering angle a 56. In a slip power recovery scheme of 3-phase
(c) By adjusting the triggering angle a only induction motor drives, what is the operating
(d) By adjusting the triggering angle a followed speed with respect to the synchronous speed of the
by armature connection reversal motor, if the power is injected into the rotor circuit
53. Ans: (d) from an external source?
Sol: 3-f full converter fed seperately excited DC motor (a) motor speed is always below the synchronous
quadrant-III operation. speed
Va → -ve, ia → -ve (b) motor speed reduce to zero
So, we have to adjusting the triggering angle (a) (c) motor speed is above the synchronous speed
followed by armature connection reversal. (d) motor speed is equal to the synchronous speed
56. Ans: (c)
54. A lamp takes 10A at 250V and emits 16000 Sol: In slip-power recover scheme motor speed is
Lumens. Determine its Mean Spherical Candle always above the synchronous speed Nr > Ns.
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63. If u(t) denotes the unit step function, which of the (b) reduced efficiency
following is an example of a bounded signal? (c) reduction in rotor slip
(a) etsin(t)u(t) (b) tu(t) (d) increased peak torque capability
(c) e u(t)
2t
(d) e-2tu(t) 65. Ans: (b)
63. Ans: (d) KSV 2
Sol: If the signal amplitude is bounded throughout the Sol: Tem =
R2
time interval −∞ to ∞, then the signal is called SV2 = constant
bounded signal. 1
So, options (a), (b) and (c) are unbounded signals, S↑∝
V2 .
option (d) is example of a bounded signal. and S↑ ⇒ Nr↓
V.
64. Consider the following lists regarding speed control fe ∝
f
methods for dc motors: I2 ∝ S↑
A. Armature voltage
I 22 R ↑
B. Field current control
h↓
C. Use of diverter resistor
D. Rheostatic voltage
66. In a transformer, the load current is kept constant,
1. Poor efficiency while the power factor is varied. Under this
2. Speed below based speed situation, zero voltage regulation will be observed
3. Speed above base speed (a) at power factor equal to unity
4. DC series motor control (b) independent of load power factor
Choose the option in which all the items are (c) load power factor is leading
correctly matched (d) load power factor is lagging
(a) A−4, B−2, C−1, D−3 66. Ans: (c)
(b) A−2, B−1, C−3, D−4
Sol: Zero regulation always possible at leading power-
(c) A−2, B−3, C−4, D−1
factor loads.
(d) A−2, B−3, C−1, D−4
64. Ans: (c)
67. The minimum value of anode current in an SCR
that is required to sustain conduction in a thyristor
65. Reducing the speed in a three-phase induction motor
with zero gate current is called
by using stator voltage control while supplying a
(a) Base current
constant torque load would result in
(b) Fundamental current
(a) higher airgap flux within the machine
(c) Latching current
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10W 10W
10W
⇒Rin =(10+10)||(10+10)
I=0 = 10W
Power dissipated in
Power dissipated in
10W 10W
the load
the load
(a) (b)
Bridge is balanced
10 × 10 = 10 × 10 Load Resistance Load Resistance
Power dissipated in
Power dissipated in
the load
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8A
Power across the load vs load resistance x
5
the load
H T H1 H 2
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frequency in rad/sec is
(a) π/4 (b) 2π R
(c) π (b) π/2 R-L
Y
76. Ans: (c) Load
Sol: Magnitude gc 1 B
0.25
e 1 3-f R-L
gc
Supply 3-f
Load
& gc rad/ sec
gc 1
For R-L load, Z = |Z|∠f
VR Vm sin t
77. Gauss-Siedel technique is commonly used in power IR = = Iˍmsin(wt - f)A
Z Z+
systems for which of the following?
IY = Imsin(wt - 120° - f) A
(a) Unit Commitment
(b) Stability Analysis IB = Imsin(wt +120° - f) A
(c) Load flow Analysis Total instantaneous power
(b) Fault Analysis P = PR + P Y + P B
77. Ans: (c) P = VmImsinwt sin(wt-f) + VmImsin (wt-120°)
Sol: Gauss-siedel technique is one of the load flow sin(wt-120°-f)+VmImsin(wt+120°)sin(wt+120°-f)
method = 3VmImcosf → constant
78. A balanced three-phase supply feeds power to 79. In the circuit below, find the terminals of X and Y,
a balanced three-phase R-L load. Under this for the circuit to be in negative feedback.
condition, the total instantaneous power supplied to 1 kΩ
1 kΩ
the load would be
(a) Pulsating with non-zero average X
(b) Zero V0
2 kΩ Y
(c) Constant
(b) Pulsating with zero average 1 kΩ
2V +
−
78. Ans: (c)
Sol: VR = Vmsinwt
VY = Vmsin(wt -120°) (a) Any of the terminals can be negative or positive
VB = Vmsin(wt + 120°) (b) Circuit will never be in negative feedback
(c) X negative and Y positive
(d) X positive and Y negative
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as per the requirement of −ve feedback the from circuit Vgs = Voltage across R
output feedback quantity which is fed back to = gm Vgs × R
−ve terminal should be more than that of +ve gm R = 1 ⇒ R = 1/gm
terminal so X should be +ve & Y should be
82. Which of the following load offers characteristic
−ve.
close to a constant load torque?
(a) Fan type of load
80. What is the hexadecimal conversion of this binary
(b) Water pumping load
number 1111?
(c) Traction load
(a) A (b) F
(d) Low speed hoist
(c) 4 (d) 8
82. Ans: (d)
80. Ans: (b)
Sol: The constant load torque applications are
Sol: (1111)2 = (F)16
• working motor have each mechanical nature of
work like cutting, grinding (or) sharing
81. For the two circuits to be equivalent, R should be
• Similarly cranes during the hoisting.
equal to
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83. Which of the following is used in overhead power armature terminals are connected to a resistance
supply for AC electric locomotive in India? (b) The voltage supply/source is reversed keeping
(a) 415V, Three phase armature terminal fixed
(b) 25kV, Single phase (c) The armature terminals are reversed and the
(c) 110kV, Single phase voltage supply/source is present
(d) 330 kV, Single phase (d) The voltage supply/source is removed and the
83. Ans: (b) armature terminals are shorted
Sol: The system of traction employing 25 kV, 50 Hz 85. Ans: (a)
1-f, AC supply has adopted by Indian Railways. Sol: During dynamic braking, the armature is removed
from source and connected to a resistance.
84. Which of the following statements are entirely true
regarding Eddy current? 86. What is the time period of oscillation of Vout?
(a) Eddy current loss can be minimized by thin R3
laminate core and Eddy current is proportional
to the flux frequency
− Vout
(b) Eddy current loss can be minimized by using
material which have low hysteresis coefficient +
C
and Eddy current is proportional to the square
of the flux frequency R2
R1
(c) Direction of Eddy current can be found by
Lenz’s law and Eddy current is proportional to
(a) 2 R1 R3 C ln e1 + R 1 o
R 2R
the square of the flux frequency
2 2
(d) Eddy current loss can be minimized by using
(b) 2 R2 R3 C ln e1 + R 2 o
material which have low hysteresis coefficient R 2R
1 1
and Eddy current is proportional to the flux
(c) 2R3 C ln e1 + R 1 o
frequency 2R
84. Ans: (a)
2
85. Which of the following statements is true in the 86. Ans: (c)
case of dynamic braking of separately excited DC Sol: The given circuit is an Astable Multivibrator.
motor?
(a) The voltage supply/source is removed and the
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R3
and relative permittivity. Wheatstone bridge is
measured medium resistance. Kelvin bridge is
−
V0 measured low resistance.
+
C
88. Find the type of filter shown below.
1kW
R2
R1
1pF
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Vi ] s g Va ] s g Va ] s g V0 ] s g
0−4 0 − Vx
R R + =0
1 SCR 6k 3k
Vx 4
Vi(s) − Va(s) = (1 + SCR) Va(s) − V0(s). (1+SCR)
3k 6k
− 12
V0(s) (1+ SCR) = (2 + SCR) Va(s) − Vi(s) Vx = ⇒ Vx = −2V
6
= 2Va(s) + SCR Va(s) − Vi(s) Ix = − 2V/3k
SCR
= Vi(s) Ix = − 0.67 mA
2
V0 ] s g 1 SCR 90. Which of the following could be the effect of using
d n ← HPF
Vi ] s g 2 1 SCR
high-speed circuit breakers in a power system?
(a) Increased short circuit current
89. For the circuit below, find the value of current Ix. (b) Reduced system reliability
(c) Improved system stability
(d) Reduced short circuit current
90. Ans: (c)
Sol: To improve the stability of the system high speed
3kW Ix 3kW
and auto reclosing circuit breakers are used.
3kW
6kW
- 91. An ideal diode connected in series with a pure
V0
4V ± + inductance is supplied from an ideal AC voltage
source. Then for what duration in radians will
the diode conduct, with respect to the AC voltage
waveform?
(a) −0.67 mA (b) -0.33mA (a) p/2 (b) p/4
(c) 0.33mA (d) 0.67mA (c) 2p (d) p
89. Ans: (a) Vx 91. Ans: (c) D
Sol: Sol:
3kW Ix 3kW i0
+
3kW
6kW
vs(t)=Vmsinwt ~ L
-
-
V0
4V ± +
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Conduction angle of diode (D) = gD = 2p-0 = 2p For stability k + 10 > 0 and (k + 4) < 0
k > − 10 and k < − 4
92. The impedance of a three phase transmission line in Condition for stability
ohms is given as Zline = 5 + j10. If the line delivers −10 < K < − 4
100 MVA of power at 400 kV, what would be the
transmission power loss in the line? 94. Consider the following statements about the
(a) 356 kW (b) 104 kW armature mmf wave in a dc machine
(c) 621 kW (d) 210 kW A. The mmf waveform has a sinusoidal shape.
92. Ans: (b) B. The mmf waveform has a triangular shape.
Sol: S = 3 VL . IL C. The mmf waveform rotates with respect to the
100 # 10 6 armature
IL = D. The armature waveform rotates wtih respect to
3 # 400 # 10 3
the stator.
PLoss = I 2L . R
Which of the above statements is/are true?
= f p ×5
2
100 # 10 6 (a) B and C only (b) B and D only
3 # 400 # 10 3
(c) A and C only (d) A and D only
PLoss/ph = 104.17 kW/ph 94. Ans: (a)
PLoss(T) = 312.51 kW Sol: The armature mmf is triangular and rotates w.r.t
armature.
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95. In a grid-connected synchronous generator working (a) Forced commutation of the SCR
at unity power factor, increasing the field excitation (b) Limiting the di/dt through the SCR
has the effect of (c) Triggering the SCR
(a) Increasing both active power and reactive (d) Preventing over voltages across the SCR
power supplied to the grid 97. Ans: (d)
(b) Increasing the operating frequency of the grid. Sol: R-C snubber circuit connected in parallel to an SCR
(c) Increasing only the active power supplied to dv
to protect the SCR against high protection and
the grid dt
(d) Increasing only the reactive power supplied to avoiding the false triggering.
the grid
95. Ans: (d) 98. A parallel plate capacitor is made with three
Sol: A synchronous generator connected to the grid dielectrics placed between two metal electrodes.
by changing excitation with constant primemover
input results only change in its reactive power, but
no change in its frequency and active power.
Material 1
(a) Both source and load have large inductances The thickness of Material 1, Material 2 and Material
(b) Both source and load have small inductances 3 is d1, d2, d3 respectively. The dielectric constant
(c) Source has a large inductance and the load has for Material 1, Material 2 and Material 3 is e1, e2, e3
a small inductance respectively. The capacitance per unit area of this
(d) Source has a small inductance and load system is given by
inductance is large
(a) C = 1 3 F
-1
d d
96. Ans: (d) 1 3
Sol: Voltage source inverters (VSI) are suitable for d d
inductive loads. (b) C 1 3
1 3
d d d
(d) C 1 2 3
97. Which of the following is the function of an R-C 1 2 3
98. Ans: (c)
snubber circuit connected in parallel to an SCR?
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Z= = G
10 2 1
Sol: As dielectric interface is parallel to metal − 105 103
plates, this configuration is series combination V1 = 102I1 + (1)I2
of 3 capacitors. V2 = -105I1 + 103I2
1 1 1 1
-I2 × 1k = -105I1 + 103I2
C eq C1 C 2 C3
1 1 1 1 105I1 = 2 × 103I2 ⇒ I2 = 50I1
1 A 2 A 3 A
d n d n d n
C eq 100I 2
V1 = 100I1 + I2 = + I 2 = 3I2
d1 d2 d3 50
− V2
1
c 1 2 3m
d d d 1 V1 = 3 d n
C eq 1 2 3 A 1000
V2 1000
C eq A ;
d1 d 2 d3 1
E Voltage gain (GV) =
1 2 3 V1 3
Capacitance per unit area is 100. Consider a circuit shown below with DC supply
; E
C eq 1
C' eq
d1 d 2 d3
(VS = 5V). The series resistance (RS) of 5W is
A 1 2 3 connected as shown. Maximum power dissipated
in the RL is
99. A two port network has the following impedance
Vs ± 5V RL
Vs ~ V1 N V2 1kW
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Vs2 (5) 2
Pmax = = = 1.25 Watts 103. The maximum efficiency of a single phase
4Rs 4 # 5
transformer operating at unity power factor is found
101. The Laplace transform of the differential equation to be 90% under full load conditions. The efficiency
y″ + ay′ + by = f(t). Assume that y(0) = 5, at half load at the same power factor would be
y′(0) = 10,Y(s) and F(s) are the Laplace transforms (a) 88.3% (b) 90%
of y(t) and f(t) respectively (c) 84.5% (d) 87.8%
(a) s2Y + 5s + 10 + a(sY + 5) + bY = R(s) 103. Ans: (d)
(b) s2Y + 10s + 5 + a(sY + 10) + bY = R(s) Sol: Assume transformer rating = 1 kVA
(c) s2Y - 5s - 10 + a(sY - 5) + bY = R(s)
1#1#1
(d) s2Y - 10s - 5 + a(sY - 10) + bY = R(s) 0.9 = ⇒ 2Wi = 0.11
1 # 1 # 1 + 2Wi
101. Ans: (c)
Sol: y″+ay′+by = f(t) Wcu = Wi = 0.055 pu.
apply Laplace transforms on both sides 1
#1#1
2
s2 Y(s) − sy(0) − y′(0) +a(sY(s)−y(0))+bY(s) = F(s) η1/2full = # 100
1 1
s2 Y(s) −5s − 10 +a(sY(s)−5) + bY(s) = F(s). # 1 # 1 + 0.055 + # 0.055
2 4
= 87.8%
102. Consider the following devices and their
characteristics
104. In a single-phase SCR based full-converter with
A. Schottky Diode
continuous conduction operating with firing angle
B. Silicon Controlled Rectifier
a, what is the angle duration of conduction for each
C. IGBT
pair of SCRs in radians ?
D. BJT
(a) p (b) a
1. Current controlled turn-on and turn-off
(c) p - a (d) p + a
2. Majority carrier device
104. Ans: (a)
3. Voltage controlled turn-on and turn-off
Sol: 1-f full converter
4. Four layer device structure
Highly inductive load
Choose the option in which all the items are
T1, T2 conducts for a to p+ a and T3, T4 conducts
correctly matched.
for p + a to 2p + a.
(a) A-4, B-2, C-1, D-3
]T1 /T2g ( )
(b) A-2, B-1, C-3, D-4
(c) A-1, B-2, C-3, D-4 (T3 /T4) (2 ) ( )
(d) A-2, B-4, C-3, D-1 Each pair of SCR conduct for p rad
102. Ans: (d)
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105. Consider a linear system represented in state space M c # ^ 3h 0 ! 0 & Controllable
form as shown below: Observability:- Mo = [CT AT CT]
xo > Hx > Hu
0 1 1
0 3 1
Now AT. CT = = G= G = G
0
3 6 0
1 6 0 1
y = [1 0] x
Mo = = G
1 0
Which of the following is true for this system? 0 1
(a) The system is stable, controllable and
observable M o 1 # 1 0 1 ! 0 & Observable
(b) The system is stable and observable, but not
controllable 106. A three-phase induction motor rated at 15hp, 440V
(c) The system is stable and controllable but not has an efficiency of 85% and operates at a power
observable factor of 0.9 (lag), while delivering rated output
(d) The system is controllable and observable, but power. What would be the reactive power drawn by
unstable the motor under this condition ?
105. Ans: (a) (a) 8.21kVAr (b) 5.11kVAr
(c) 7.42kVAr (d) 6.38kVAr
Sol: Given A = G, B = G , C = [1 0]
0 1 1
3 6 0 106. Ans: (d)
C.E. ⇒ SI A 0 Sol: h =
output 15 # 746
=
input 0.85
[SI − A] = S = G − A
1 0
0 1 ⇒ 3 # 440 # I 2 # 0.9 = 13164.7
s 1
G= G
I2 = 19.19 A
[SI A] = G =
s 0 0 1
0 s
3 6 3 s6 ∴ Q = 3 VI sin
s 1 = 3 # 440 # 19.19 # 0.436 = 6376.4 VAR
C.E. & 0
3 s6
C.E. ⇒ s (s + 6) + 3 = 0 107. A dc shunt motor supplied from a 220 V DC input
C.E. ⇒ s2 + 6s + 3 = 0 supply runs at 1200 rpm. Neglecting all losses and
Roots are Negative, lies in the Left half of s-plane saturation, what would be the speed when the same
hence given system is stable. motor is supplied from a 175 V DC input?
Controllability: ⇒ Mc = [B AB] (a) 1200 rpm (b) 1100rpm
(c) 750 rpm (d) 950 rpm
Now A.B. = = G= G = G
0 1 1 0
3 6 0 3 107. Ans: (a)
Sol: The speed of shunt motor is almost same with
Mc = = G
1 0
0 −3 change in voltage.
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108. 4 mA current flows through a conductor for 4s. 110. A bipolar junction transistor is
Find the number of electrons passed through the (a) a charge-controlled device
conductor. (b) a field-controlled device
(a) 1e14 (b) 16e14 (c) a voltage-controlled device
(c) 16e-6 (d) 16 (d) a current-controlled device
108. Ans: (a)
Q 110. Ans: (d)
Sol: I=
t Sol: In a BJT, input current controls the output
I=
ne current, hence it is considered as a current -
t
controlled device.
I#t
n= e
4 # 10 −6 # 4 111. What is the way of imparting braking action of
=
1.6 # 10 −19 three phase induction motor in plugging scheme?
∴n = 1×1014 (a) By reversal of the phase sequence of the stator
(b) By increasing the stator voltage magnitude
109. Two 10-bit ADCs, one of successive approximation (c) By decreasing the stator voltage magnitude
type and other of single slope integrating type, take (d) By decreasing synchronous speed
Ta and Tb time respectively to convert 3V analog
input signal to digital output. If the input analog 111. Ans: (a)
signal is increased to 6V, the approximate time Sol: Plugging is possible by reversal of phase sequence
taken by the two ADCs will respectively be of stator supply
(a) 2Ta, Tb (b) 2Ta, 2Tb
(c) Ta, Tb (d) Ta, 2Tb 112. Which of the following statements is correct in the
109. Ans: (d) case of multi-loop based close-loop control of DC
Sol: In successive approx-type ADC, Tconv = n.TC to the and AC drives having speed and current feedbacks?
conversion time is independent on input voltage. (a) Current control loop is faster than the speed
• In a single slope ADC, the conversion time is control loop
proportional to input analog sample magni- (b) Output of current control provides the
tude and it is having linear relationship hence reference speed
when Va is changing from 3 V to 6 V (c) Speed control loop is faster than the current
Ta becomes Ta (in successive ADC) control loop
Tb becomes 2Tb (in single slope ADC) (d) Speed control and current control loops have
equal bandwidth.
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114. What is the status of machine flux, for the range of 10kΩ 5kΩ
frequency above the rated (base) frequency, in the V0
case of a speed regulated V/f controlled induction
Vin
motor drive? g=∞
(a) Machine flux unchanged
∞
(b) Machine flux may increase or decrease based
on load 10kΩ 1kΩ
(c) Machine flux increases from the rated flux
(d) Machine flux decreases from the rated flux
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The given circuit is common-source (CS) amplifier 118. An electric machine wherein the self-inductances
with unbypassed Rs. of both stator and rotor windings are independent
The formula for small - signal gain in such circuit of the rotor position will NOT develop any
gm RD (a) Hysteresis torque
is, A V = − 1 + g R ........... (1)
m S (b) Synchronizing torque
gm RD (c) Starting torque
AV = − ............. (2)
g m RS (d) Reluctance torque
[a g m = 3 & g m RS 22 1]
118. Ans: (d)
R
A V = − RD
S
119. If the Nyquist plot of the open loop transfer
5k function G(s)H(s) of a system passes through the
A V = 1k 5 .......... (3)
-1 + j0 point, then the phase margin of this system
116. The following gates are designated as Universal is likely to be
Gates (a) 90 degrees (b) 45 degrees
(a) NOR and NAND (c) 180 degrees (d) 0 degree
(b) NOT, OR and AND 119. Ans: (d)
(c) XOR, OR and AND Sol: If system passes through (−1 + j0) then system is
(d) XNOR, NOR and NAND marginal stable [M.S.]
116. Ans: (a) For M.S. ⇒ G.M. = 0 dB and P.M.= 0o
Sol: NAND & NOR gates are universal gates.
120. At low values of operating slip, the torque
developed in a three-phase induction motor is
117. In a three-phase bridge inverter operating in the
(a) inversely proportional to slip
square wave mode, the output voltage waveform
(b) proportional to the square of slip
contains
(c) linearly proportional to slip
(a) Only even order harmonics
(d) independent of slip
(b) Both even and odd order harmonics, but no
120. Ans: (c)
triplen harmonics
Sol: For low value of slip region, Tem ∝ S.
(c) Only triplen order harmonics
(d) Only odd order harmonics 121. The pole zero plot shown below represents a system
117. Ans: (d) whose frequency response is approximately that of
Sol: In 180° square waveform is symmetric then all even a
harmonics are absent. Only odd order harmonics
are present.
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∴ L′ = 16.1mH/m
w (in rad/sec) 124. A set of 4-point charges of 3mC are placed at (1,
So, it is Band-pass filter 1, 0), (−1, 1, 0), (−1, −1, 0) and (1, −1, 0) in a
space with relative permittivity of 2. Calculate the
122. If Pe and Ph denote the eddy current loss and resultant electric field at the (1, 1, 1).
hysteresis loss in a magnetic core operating with an (a) E =16.4 xt + 16.4 yt + 3.41 zt kV/m
alternating flux density waveform of frequency f, (b) E =−16.4 xt −16.4 yt + 3.41 zt kV/m
then (c) E =−3.41 xt −3.4l yt +16.4 zt kV/m
(a) Pe and Ph are both proportional to f2 (d) E =3.41 xt + 3.4l yt + 16.4 zt kV/m
(b) The total loss (Pe + Ph) is proportional to f 124. Ans: (d)
(c) Pe is proportional to f and Ph is proportional to f2
Sol: E E1 E 2 E3 E 4
(d) Pe is proportional to f2 and Ph is proportional to f
1 ^2at x at zh
RS V
122. Ans: (d) at z 1 WWW
e o
SS 1
9 # 109 # 3 # 10 6 SSS ]1 g ]1 g ^ 5h ^ 5h ]3 g2 WW
S W
Sol: Ph ∝ f
2 2
T X
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RS V
2at x 2at y at z WW e L _e L − 1 i
SSt 2at x at z V −Rt −Rt
(b) i(t) =
o
Sa z + + + + + W
27 27 WWW R
27 # 103 SS 5 5 5 5 27
= SS WW
e L _e L − 1 i
V −Rt Rt
2 SS+ 2 at + 1 at WW (c) i(t) =
o
SS y z
WW R
5 5 5 5
e L _e L − 1 i
T X V −Rt −Rt
(d) i(t) =
o
RS 2 VW
SSf +
2
p t + f
2
+
2
p t WW R
a a
# 103 SSS 5 5 27 WW
x y
= 27 27 5 5
WW V/m. 126. Ans: (c)
SS WW L
2 SS+f1 + 1 + 1 + 1 p at z WW Sol:
SS 5 5 27 5 5 W
T X ± V0u(t)
∴ E 3.41xt 3.41yt 16.4zkV
t /m . Vs R
- V u(t-t )
+ 0 0
i1 L L
S domain
Given that f(x) is an even function. Which of the
following option is correct ? V0 R
± s
(a) bi = 0 (b) k ≠ 0, ai ≠ 0, bi ≠ 0 I(s) SL
(c) k =0 (d) ai = 0 -
+ V0 −st0
s e
125. Ans: (a)
Sol: For an even periodic signal bi = 0 ∀ i.
V0 V0 −st0
−
s s e = I(s)(R + SL)
126. Consider a series RL circuit shown below RS 1 VW
V0 (1 e ) V0 SS WW
st0
L
SS R WW^1 e st0h
L SS s cs m WW
I(s) = s
(R sL)
L
± V0u(t) RS1 VW T X
1
R V0 SS WW
- V u(t-t ) I(s) = Ss R W^1 e st0h
+ 0 0 R SSS s WWW
L
TR X VW
SS1 1 st0 1 1 st0W
V0 S s s e e WW
u(t) and u(t − t0) are unit step functions. The current S
cs m cs m
I(s) =
R SSS WW
R R
flowing through the resistance R at time t > 0 is L L W
T X
` e L t e L ^t t 0 hj
V0 R R
given by By I.LT, i(t) =
R
e L _e L − 1 i
V Rt −Rt
(a) i(t) = e L _e L t 0 − 1 i
o
V0 −R t R
R i(t) =
R
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127. Calculate the amount of the point charge at the 129. For the parallel operation of 2 single-phase
origin given that the potential at (−2, 3, 1) is 36 V transformers with same voltage ratio and different
and reference is taken to be at infinity. kVA ratings, the load is shared by these transformers
(a) Q = 7.5 nC (b) Q =1.5 nC in proportion to their kVA ratings when the
(c) Q =15 nC (d) Q = 30 nC transformers have
127. Ans: (c) (a) leakage reactance in ohms inversely
Q proportional to their ratings.
Sol: V (b) equal per unit impedances on their respective
4R
ratings.
R ] 2g2 ]3 g2 1 2 14
(c) the same leakage reactance in ohms.
Given V = 36Volt (d) the same magnetising reactance in ohms.
9 # 109 Q 129. Ans: (b)
36 = Sol: The two transformer’s should have equal pu
14
impedance based on their respective kVA ratings.
36 # 14
Q=
9 # 109 130. A three phase induction motor with a 6-pole winding
∴Q ≃ 15nC is rotating at 1200 rpm. The speed of rotation in
electrical and mechanical radians per second are
128. A step up chopper delivers an average output respectively
voltage of 100 V from an input supply of 60 V when (a) 40π, 40π/3 (b) 40π/3, 40π
operating with a continuous source current. What is (c) 120π, 40π (d) 40π, 120π
the operating duty ratio for the switch ? 130. Ans: (c)
(a) 2/3 (b) 1/3 2N
Sol: wm = (mech)
60
(c) 0.6 (d) 0.4
2# 1200
128. Ans: (d) = = 40p
60
Sol: Step up chopper P
we = m
2
V0 = 100 V, Vs = 60 V 6
D=? = # 40
2
Vs
V0 = = 120p
(1 − D)
60
100 =
(1 − D) 131. What is the SNR of an ideal 10 bit ADC ?
(a) 71.96 dB (b) 81.96 dB
D = 0.4
(c) 51.96 dB (d) 61.96 dB
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131. Ans: (d) 134. A 200 Volt, 1200 RPM, 100 Amp DC separately
Sol: For an ideal n - bit ADC SNR is 6.02 × n + 1.76 dB. excited motor has an armature resistance of 0.1Ω.
For n = 10, SNR = 61.96 dB. It is braked by plugging from initial speed of 1200
RPM. What is the external resistance to be placed in
132. A motor-load combination operating in motoring series with the armature circuit to limit the braking
mode (quadrant-1) has the following speed torque current to twice the full load value ?
characteristics : (a) 10.2 Ω (b) 7.33 Ω
T(motor) = 0.1 N - 10 (N-m) (c) 1.5 Ω (d) 3.8 Ω
T(load) = 0.25 N - 75 (N-m) 134. Ans: (*)
where T(motor) is a motor torque in N-m, Sol: Eb = V − IaRa = 200 − 100 × 0.1 = 190 V
T(load) is a load torque in N-m and N is a speed of V Eb 200 190
the motor-load combination in RPM. What is the 2Ia =
Ra R B 0.1 R B
steady state speed of the system after exciting it ? ⇒ (200) (0.1 + RB) = 390
(a) 700 RPM (b) 666.3 RPM ⇒ 20 + 200 RB = 390
(c) 433.3 RPM (d) 0 RPM RB = 1.85Ω
132. Ans: (c)
Sol: TM = 0.1 N -10 135. A 200 Volt, 1000 RPM, 100 Amp separately excited
TL = 0.25N - 75 dc motor has an armature resistance of 0.1 Ω. The
N=? motor is fed from a DC-DC step down chopper.
At steady state The input dc source has a voltage of 300 Volt to
TM = TL this chopper. What is the duty cycle of the chopper
0.1N -10 = 0.25N - 75 for motoring operation at rated torque and speed
∴ 65 = 0.15 N of 500 RPM assuming operation with continuous
65 conduction and field flux remains unchanged ?
∴N= = 433.33 rpm
0.15 (a) 0.50 (b) 0.35
133. Which type of transformers is used in AC welding ? (c) 0.25 (d) 0.84
(a) Step down type (b) Equal turns ratio type 135. Ans: (b)
(c) Ferrite core type (d) Step up type Sol: Seperately excited dc motor,
133. Ans: (a) Va rated = 200 V, Nrated = 1000 rpm
Sol: The type of transformer used in the AC welding set ia rated = 100 A, Ra = 0.1 W
is step down transformer. It will give low voltage Step down chopper
and high current on secondary. Vs = 300 V, D = ?
T2 = Trated, N2 = 500 rpm
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139. Initially the switch is closed, and steady state has 140. Ans: (b)
been reached. At t = 0 the switch is opened. What Sol: No energy will be stored (or) loss in magnetic
is the voltage across the capacitor at t = 0+ ? material.
R 4R
141. Consider a circuit below with Rp = 5Ω.
C 5R t=0
±
- 60V
I 5A RP
(a) 0 V (b) 60 V
(c) 10 V (d) 50 V
139. Ans: (d) It is equivalent to which one of the following?
Sol: RS
(a)
2
R 4R
C 5R t=0 VS R 2
S RS =5W ±
± 2
5V
- 60V
(b) RS 3
-
RS 2
60 # 5R (d)
By VDR, VC(0-) = = 50 Volts VS R
5R + R 1
RS2 =5W ±
VC(0+) = VC(0-) = 50 Volts 25V
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144. Find the polarity of op-amp input for negative ⇒ If A is positive then only the loop gain can
feedback operation. be negative
‘A’ should be positive, ‘B’ should be negative
VCC
I C R L
1
(a) The circuit will never operate in negative (a) ω0 = Hz
LC
feedback 1 R 1
(b) The circuit will always operate in negative (b) ω0 = + + Hz
RC L LC
feedback irrespective of the op-amp input 1
polarity (c) ω0 = Hz
RC
(c) A is positive and B is negative R
(d) ω0 = Hz
(d) A is negative and B is positive L
144. Ans: (c)
Sol: +VCC 145. Ans: (a)
I = 1 mA Sol:
I C R L
A
1V B
1
Resonance frequency, 0 Hz
LC
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1 b x x l
2 r
2 2 -r
ln d n
2 0.68
n ln
n1 a 0.1045
/ b 2n sin ]ngl cos(nx)
3
(b) + 0.0169 # 10 −9
2 n1 =
/ b 2n cos ]ngl sin(nx)
3 1.87
(c) p +
n1 C ≃ 9.05 × 10−12 (or) 9.05pF
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TSPSC - 2022
Telangana State Public Service Commission
Electrical Engineering
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Questions with Detailed Solutions :: 45 ::
22-01-2023
R12 2 R 2
N=0 Ns
S=1 ← S=0
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